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Written by :A SHAHZAD AHMAD KHAN ~

- ~ ‘ ' LECTURER (Mathematics) .

G.C.KOT SULTAN (LAYYAH)

l ct f A -—> R be a function. Clearly domain of f A, in other words»l Real Valued Function '

fl 4

is d<-:fi.n‘éd On A. Slinc0fc."o-clo1lnai11 of "f is R, ""~-/\/6' can say that‘ f.‘isA1rcalll

valued fifmcti011. 'X ~

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Mwiyl X -

Let X lav.‘ a inon-empty set and R be a/<rea"l uur11be1"s.l

;;- Let d_:__X_ >< X —'» be a function

M4

'-5 4 11 - /1 - 11 u - - ' " ‘ ‘ -

Then "d" _1s_callecl mgtnc on X, 1f d s;-1t1.:=f1es each of the followmg four

conditions; - - . \ ' ‘

(M1) Ld(.x'1-X2) 2 O V-Y-1»3\'2

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. / ,I - ‘fr

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The non-negative '21-:a.l nu..1nl.".~;-':-.1." d(x,,x-Z) is c21.'=led. clistancc: betw'e<2n__

‘points xi and x2 in the metric

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13_>s2i1u?.1.eLet X be the set of all towns marked on a plane geographically map

and let d(x1, x2) be the length of the shortest 1."-out L from town x1 to x2.

Show that "d" is a metric on X "S

SolutionHere function d.X >< X —> R IS defined as

‘ d(x1, x2) =Length of shortest route lfr_o])1l[tQw_n__gc1_‘c2_x2.

(M1) Since (Length of shortest route from towny to x2) 2 0

-»

d(x1,x2) 2 O-

~

v.

1

(M2) Let d(x1, x2) = => Length of shortest route from town x1 to x2 = O

-' "

=>x1=¥2 L

l

LLet x1 = x2 => Length of shortest rou{Q ljrom town x1 to 2;; = O

W Ii :> d\(x1Ix2) = O

‘ Y

, (P-/I3) Since d(x1, x2) = Length of shortest route From town x1 to x2“ _"§

\ . 4.: '‘

= Length of shortest route -from town x2 to x1

Z d(x2»x1) I

H

(M4) Let x1,x2,x3 E X

Then “x1,x2,x3 are non-collinear or collinear '

If x1,x2,x3 are non-collinear, then they forynua triangle and we know

that sum of-length of two sides of a4triangle_ 15' always greater than the

third side.‘ .

d(x1,x2) + d(x2,x3) > d(x1,x3) ----------------- (i)‘

Let x1,‘x2,x3 are collinear.

Thén d(X1, X2) + d(X2, X3) : d(X1', X3) """""" "-

From (i) and (ii), we get

l-lence d is a metric on X.W Rd<x1.x7.>+d<x2.x3>zd<x1,ix3> X

ExampleLet X = R be the set of all real number: and let d:R >< R —> R be

defined by d(x1, x2) = lxl —- xzl denotes the absolute value of the

number x1 — 'x2. Show that (R, d) is a metric spagxa.

Solution '

Here function dz R >< R ——> R is defined as

s dlxnxz) =' lx1"' xzl

(M1): Since lxl - X2‘ 2 0

d(x1,x2) 2 O ‘/ '

(M2) Let d(x1,x2) = O => \x1g— >52] = O.

=> x]-x2=0:> X1:

cg

LQL X1=X2 =>'X1#-X2:" ' 2 \X1;X‘2\ 0 * '

Thus’ d(x1.'x2) = O <=>" x1 == X2

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(M3) Since d(X1:X2)=\X1'X2\ _

%. /;g.-/ 4

Exarn1;1_§

= X3-_-X1

= d(X2»X1)

=|"(X2 '."X1)|

M4,) _ Si1’1C€ d(X1,X2) = X1v

d(Xz_,X3) = \X2__' X3

am.-x3) = x1 — x3‘

NOW 'd(X1'_X3)= X1“X31 '

‘ .. ;. i‘

' " ' = dx1v—~2<;+"X2+X3l_ ‘

. .r.—""*”‘ . "“""'_’_ -

<, X1 =1 Xi; _+ \X2. '.»'X3l

'-—_ d(X{'.,.x.2) + d(x21x;)- __

" VThus(R; d) is a metric space. A" '

.

‘W _

,

- Let X=R2 be a set of all ordered pairs(x,y); -x,y JGR. Let

P1(x1,y1),P2(x2,y2) e R2. Show -that the non-negative real -valued. V.

lutmn.

X Here function dz R2 >< R2 —> R is defined

> an)1,P2):lX1 "X2\ + W1 *3/21

(M1) /Si11Ce |X1“" X21 + W1 Y2‘ 0 r

d(P1,P2) _{ O-Q-..

8

(M2)- La d(P1iP2):0 :>\X1"X2\‘+\}/1"‘)/2\=O=1’ \X1"X2|=0, U’ *“ =

1

,=> ex

1' 3

\

I2 P-L 7: P2 '

_____,

1 y2‘ O5

1 “ X2 = 0, 3'1 " Y2 = X

/ 4 I

Let“ P1 = P2 => (X1»}’1)=(X2-Y2)I

I

‘

=> X1=X21 3'1-“M1V

- Oll

=> X1 “Xx =' 0,-_ Y1 "Y1? 4‘| ' | ' n

=* \X1 -12! 0- £1/~. ‘/-¢1.=-1

Ir _\

Z? '

Thus 'd(P1,P2) = O ¢=> P-_, = P2

(M3) since d<P1,P2>=\x1—x2!+mwzl

r

I (M4) Si-‘TC? ‘ CKP1, P2) '3 ‘X1; '- 3:721 ix)’: " .3/I21 I

1

' = ]"(X2 X1): + i"(3'2 " )’1)| .

='-' “ X1!'+ ‘Y2 '" 3'1 X

- 1

= dll. P1)

1

furretion "d" defined by d(P1,iP2)‘ = lxl F x2\-+ iylf-~yg1 is a metric on R3. ‘L

a

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=1’ X1=X2' }’1:)’2;A

(X1»V1) '1 (X2»}’2) »~

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I

:> X- "X": "§' E)/1 7:

d(P2»P3) : 1X2 "' X14‘ *1‘ W1 ” }’::\2

d(P1»:1X1.," + 1,‘/1 " Y‘/".~r5

\' '

-

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4

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‘ _ — """""_"£T'.__ :-_»=:=~__E--_7. ._-~».-==;-.;=.=._-.—.;:4;;~;.—;==_»'¢<¢--1:- *1 .._, ” '_‘ " "" ‘“ "' ' '

— - - - —- ---- »- ~-» _ - - -.1 .. r. M314 _ ,.,;—§£-.=w1-:'.'a_¢.‘.-;-.—:--r::1r;s>r;_:-;;‘ ~~—»~---

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I (M3) Since d(p1» P2) : max (‘xl ‘ X2‘ 1 ‘Y1 " 5'2‘)

i‘

Since d(P1,P3) = \x1— x3‘ + \y1 — y3\

w=‘Xi“x2+x2"'X3""‘)"1"')’2+}’2")’3‘5 ‘X1 ‘ X2‘ "' ‘X2 "X3‘ + ‘Y1 —)’2‘ + ‘Y2 _)’3‘

=\x1 —><2i +ly1 —y2\ + \- 1—X3\ + W2 “)’3‘

= d(P1,P2) + d(P2,P3)

Hence "d" is metric on R2. _

~

ExampleLet X=.R2 be a set of all ordered pairs(X.)’); x,yeR. Let

P1(x1,y1),P'z(x2,y2) ER2. Show that the n01--negative real valued

function"d" defined by d(P1, P2) = max (‘xi x;| ’ lyl - yzi) is a metric

on R2. »

R

S01uti9_n_

"Here function dz R2 >< R2 —> R is defined as

d(P1»P2) = max (‘X1 “ X2‘ + ‘Y1 " Y2’)

(M1) Since max (‘X1 ‘ X2‘ “F ‘Y1 “ Y2‘) Z O

(""x1"'x2‘Z0&‘)'1“)’2‘?-O)‘

d(P1,P2) 2 O

(M2) Let d(P1»P2)=0 => m_aX(‘X1-x2"‘y1".Y2‘)=0

=> lxl-><21=o. w.-y21=0 I

:> x1——x2=0, —-y2=0

<1

rri

¢.

v

I

E

»4-Liz.-=;.;—_:~,|_=_,_-.__.-,..._\._»._‘._J._

\

1

k

2* x1:x21 T-Y2 x

._./\\<

=>_ (X1,Y1)=(x2-Y2, -

P1:-‘P2

7'Let P1 = P2 ;’ (xvyi) = (x2:}’2)

““ " =9 x1=x2- }’1=Js-.

‘ => max (‘x1 “x2‘9‘)'1 ” Y2‘) = O‘ -

:> d(P1»P2)=O .

I ThLlS‘ d(P1,P2)-T-O ‘ti P1-7P2

U

3?>-\

R

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Z5

@

= max<\—<x2 — ><1>1,+1-- 1» —y1>\>

= m(lX(‘x2 “ x1‘]' ‘Y2 "' M‘)

= d(P2» P1)

U

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CD

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Xd(P1.P2)=.[(X1—X2)2+(y1-312?]?

'

i

d(P1-P3) = maX(|X1 " Xsiij W1 "é}’3\) = \X1 “ X3‘ (SEW)

"""":‘r~».._

/\>-I

/

\

J

i ,,a

(M4) X since d<P1,P2> = max cm - ><2\,+\y1- mi = _\X1 -X Xzi (Say‘1

XX

d(P2.P3)=maX(\Xz-—X3l7\y2—y3\)=\X2—X3\(say)

NOW d(P1IP3):\x1 ' X3\.

X

. ' =\X1’X2+X2"X3\- X "X2\+\X2“X3\

V

:d(P1»P2)+d(P2»P3).

(We can get the same results in the remaining cases.)

Hence ”d" is metric on R2.' V

Ii\_

. 7 V’'

X

Egmk // ~”»

iet X=Ix;be a- set of all ordered pairs(x,y); _x,yeR. Let\ L

P1(x1,y71§P2(x2,§/‘gShow that the non-negative real Vvalued -

function "d" defined by d(P1,P2) = [(x1 -— x2)?‘ + (y1 — y2)2]% is a metric '

On R2."L 1 r‘

§_Q1uti0n_-_

' Here function dc R3'>< R'B'—i R is defined as ‘"M '

'‘

I’ .

(M1) ' Since [CX1 " X2)? + (Y1 " Y2)“; Z O QX

A ‘ i___,.:..

"@ ¢“ummgz0'

c

1'-.ll

' (Mzj ’- Let d(P1»P2) ='.0‘ =5 KX1 " X2)2 + (Y1 " Y2)“; = O»i ‘

X~ =-» <><1-x2>2+<y1—y2>2=0n,

X

=> (X1 ‘ X2)2 = 0. (Y1 " Y2)2 = 0

:> X1"X2=O' }’1’)’?=O 1

._:> X1=X2: )'1=)’2 V

.=> (X11)’1)=(X2»)’2)

U.

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4

1" P1=P2,

Let P1 : P2 => (X1-Y1) 1: (X2!)/2)

A=7 X1=X2»‘ )’1:)’2

=>,X1"'X2=O1 -)’1"‘}’2=0 -

=° (X1 " Z 0' (Y1 "-Y2)2A= O

\:> (X1 " XQ2 4‘ (J/1 " )’2)Z =0

=9 KX1 " X2)2 +(.)'1' )’;)2]5 =‘ 0

=> d(P1,P2) = O

i

A

r_"-1.-r,_.-_<_——_—'-==—===c

‘F *1‘ ” 1-411;;;;t‘"'":r:::_"-7-“"1 ~--“~33---'—~'-A~~—~——»~--e--~~~~- 7- -V ii , ...-___.__._,..________~_

____,1

~~ »~~~~ I»-_;__.=iff-‘*-——,"'——;j:'__*-_;";ll?""

--M W .. _:.__,, ' 1; """""

' ~———~ -~ A -~ rig- ._" "_ ' ‘f_;;:.~.:~'.-:1\;—;_‘_{§:7:'

l

I

(M3) since d<P1.P2>=t<><l—><2>2+<y1—>»2>1%4

= l{—(Xz - X1)? + {—(_>/2 "y1)}2l2

‘ t

ty.org

1

§

\

ll‘

‘ 1

I

\

I

1

I

l

>

\.

i R

A

l

= [(952 " x1)2 + (Y2 _ IV1)2

= d(P2»P1) i

(M4) Let P1(x1,y1), P2(x2,y2), P2(x3,y3) e R2_lfl'18] P1, P2, P2 are collinear

A or non-collinear.

If P1, P2, P3 are collinear, then _’

‘K "K ,1“

. d(P11 P2) + d(P2»P3) 1.‘ d(P1' P3) """""""""" (1) '

~ If P1, P2, P3 are non-collinear, then they ft :"m a triangle and we

know that, we know that sum of length of 1 -vo sides of a triangle is

always greater than the third side.

d(P1' P2) + d(P2» P3) > d(P1iP3) """""" l'“" (2) ‘

From (1) & (2) we get, .

V d(P1,_P2) + d(P2, P3) 2 d(P1,P3) "

Hence "d" is metric on R2.

bleat

www.mathc

Avaia

Example A,

,Let X=%be a set of all ordered pt ;‘€.rs(x,y); x,yéR. Let

P1(x1,y712J?,lP2(x2,)/2 Show that the non- tegative real valued

.,.,-2....‘..rv.,,1______-W.

=

t

t1

l

l

1

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,.

I

K

i

1 I

function "d" defined by" d(P1, P2) =’[(x1 — x2)2 -l- (L1/1 - y2)2]5 is a metric

- 2 'T.(,iT’2‘;)Solution »

Here function dz R? >< R%—> R 1s defined a.

d(P1»P2) = [(751 ‘ x2)2 + (Y1 _ }’;z)2+‘ (Z1 " ZzlzlgM

l ll

(M1) Sin“! K-X1 " x2)2 + (Y1 * Yzlz + (Z1 " Zzilzlii 2 O

d(P1'P2)ZO '

(M2) Let d<P1.P2> = 0 => t<x1— xm + (Y1 y >2 + <21 — z2>*1%= 0

=9 (X1 " Xzlz + (Y1 "'" Y2); + (Z1 “ Z2)2 : O

' Z‘ (X1 ” X2)2 —'= 0, (Y1 “}’2)2 Z 0» (Z1 " Z2)2 =073> x1‘_X2:0, y]":¥_:0, Z1'_Z2?0

3' X1 = X2» Y1 2‘ Y2 Z1 = Z2

=’ (X1-Y1-Z1) = (x2»l.V2-4 Q

Qrs

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‘ I

:> P1:P2‘ t

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»__1--»--»<»~_~:¢ at-__A _._ _ _,-_____._.________.._....i._."4 H ‘-4 _ _ - , . ._i.__,_i___4____,_,___,__,__,_____

L, ,

X 1 " "—*"="~' nn*~ " =‘>,,,,-“'i*"'_" ""1_"{,~,,:£-,j""=?-»<~---.=e,,»~--1--ii”: »-==-f_--j ~--—~~~~-———~—-»-—----~-J»--V------1

-’

Let-. P1 = P2 =7’ (X1-)’1»Z1)=(x'2'}’2»Z2)

lS

(M3) Since d(P11P2) = l(X1 " x2)2 + (Y1 “ )’2)2 'l‘ (Z1 " Z2)2l5 _

‘

L 75,

1.

=> X1"-'x2» 3'1 =lY2/ 'Z1=Z2 "

‘

=7’ x1"'x2=O/ )’1")’2=O/ Z1"'Z2=0

=> (x1“x2)2 = O1()’1 "')’2)2 = O|(Z1’Z2)2 = 0

2 (X1 “ X2)2 + (Y1 ' )’2)2'+ (Z1 " Z2)?‘ = O '

= Km—xQ2+Ur—wY+%a-¢Qh§=0, t

d(P1rP2) = O1

‘

2

t=u+a—anH++»-mn+++@-aW_ i

m—m%WrmYH%J»_= d(P2-P-1) ~

collinear or non-collinear.

-' - If P1, P2, P3 are collinear, then

d(P1_,P2) + d(P2»P3) = d(P1»‘P3)

(M4). ~Let P1(x11}’1-Z1):P2(x2'3’2»Z2)/PalxslyslZ3)

E lg then P11_P2» P3 are 1~-

------------~m_» /If P1, P2, P3 are _non-collinear, then they, form a triangle and we. j

know that, we know that sumgofy length of two sides of a triangle is »

_, always greater thanthe third side. .

' From (1) & (2) we get,

d4(P1»P2) + d(P2»P3) Z d(P1,P3)

Hence ”d" is metric on R5.

Example

' be converted into metric space. Y

S ' 0 SolutionLet X be any non-empty set.

ALet d,,:X >< X —> R be defined by

d°(0 ifx =x' -

"1"‘2)=l1 ifxllixiWe shall prove that do is a metric on X.

d(P1,P2) +d(P21P3) > d(P1,P3) """""" '4 '-

I

Pa,-1ae<.=1r1,s(.=.,.,,i

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(M1) l-lere dQ(x1,x2) 2 0 ('-' d@(x-,,x2) = 0 or d°(x1,x2)-= 1)

(M2) Let d,(x1,x2) = 0 =1» x, =x2 (Bydefinition) -'

Let_ x1 = x2 =§A d,,(x,,x-3) = O (By definition)

{Thus d,(x1,x2)'=O <=> xi: 1t

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Show that every non-empty set can be given a metric and hence can

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(i) Let d°(x1,x2) : O => x1 = xz (By ckefinition) ~

=> x2 = xi

d=(x2:xi) = 0

(ii) Let dO(_x1,x2y) = 1 => x1 zt x2 (B; definition)

X

R Q >=> x2:/=x1 )‘ => d,(x2,x1) = ‘.1

J L it "i;§He11ce in both the cases d°(x1,x2) = d°(; ,|,x1)

(M4) . Let x1,x2,x3 E X

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“ '81 .d°(x2,x3) =0

also d,(x1,x3) = D

*d(x1,x2) + d(x2,x3) = d(x1,x3)

(ii) Let x1 vex; at x3_ then do(x1,x2) =1_

- & d,,(x2,x3) = :1.

also d,,(x1,x3} = j_

d(x1-X2) + d(x2'x3) > d(x1.x;l=) L

. . “l' . .

Similar type of verificagionin all remaining cases 1Qi;1dS us to the conclusion

that d(x1,x2) d(x2,x3) 2 d(x1,x3) _V )(1,x2,x3 e X 7

Hence (X, do) is a metric space.

NoteLet X be any non-empty set. Let do: X >< X —+ /K be defined by

1:

‘iii '0 ifx1= - \

wen do is“ called discrete metric on X.

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t 'andletd:C><C—+Rbe- Let C be the set of allcomplex num were

defined by d(z1, Z2) = lzl -— zzl dis a metric on C

Solutioni Here function dz EX C_—> R is defined as

d(Z1'Z2):iZ1 " Z2‘

d(Z1'Z2) 2 0

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i (M2) iL9t d(Z_1»Z2) = 0 =3‘ izi “ Zzi = O‘

2 Z1"'Z2=O2 Z1322 i

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X2 , .

do(x1lx2) _' xi i x2 /

.___4.._,.-_.-__->‘_....=~'-_I~_A‘Lw_'_4...._._e......_.__-___.'

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Let 21:22 => z1—z2=0

’ => I21

Thus d(z1,z2) = 0 <=> 21 =

(M3) Since

. "_Z2l =0‘

3 d(Z1,Z2) =0

d(Z1'Z2) = Z1 " Z2. i

=1-(Z2 -Z1)! Ir""'"-~‘——*>-+-=-..._.,....,.*.,,__.____.»'_. ___

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= Z2 " Z1 Z

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.= d(Zz.Z1) ~ 3 g »

e (M4) Since» d(Z1’ Z2) : Z1 Z2 Mergjngjwan and maths .

d(Z2lZ3) = Z2 " Z3 A "M—~_#_~w-“~'“

Z

d(Z1,Z3) = Z1 _ Z3l

L__.__.

Now d(z1,z3) = 21 — 23 '

§f~

=21-—z2+’z2-+z3\ »V

' .

5 Z1_Z2.+|Z2_Z3\Z

V

: d<Z1rZ2:, + d(ZZ_I Z3)

Th (C_,d)isametric space. -

Let d be a metric on X and let d’:X >< X Z-> R be giyen by V

' d'(x1,'x2) = min (1, d(x1,x2)). Is d’ a metric on X? '

Solution" ‘

__€__.i.__i—He/re function d’: X >< —> R be given by '

d’(x1,x2) -= min (1,d(x1,x2)) X

, (M1) Since min (1,d(x1,_x2))20.

I

; -'- d,(X1,x2) Z O

I _ (M2) Let d’(x1,x2) = O => min (1,ci(x1,x2)) = 0' '

’ '=> d(x1,x2) = 0 1 ¢ 0

' => x1=x2 . I‘-'dismetric0nX. »

H. Let x1 = x2 » => d(x1,x2) £_O dis metric on X.

. . => min (1,c'i(xA-1,x2)) -= O

’ . => d'(x1|x2) = O

Thus _d’(x1,x2) = O <=> x1 = x2

(M3) Since d’(x1,x2) = min (1,d(x1,x2)) 1

' ‘ = min (1, d(x2,x1)) d is metric on X.

. ‘ = dI(x2»x1)

(M4) Since d’(x1,x2) = min (1,d(x1,x2)) = d(x1,§c2) (Say)

4

d’(x2,x3) = min (1,d(x2,x3)) = d(x2,x3) (Say)

d’(x1,x3) = min (1,d(_x1,x3)) = d(x1,x3) (Say)

Since d is a metric on X. »

d(x1/X2) + d(x2*x3) 2 d(x1/X3’)

Z’ d'(x1|X2) + d/(X21373) Z dI(X1/X3)

V We get the same result in the remaining cases.

- d’ is a metric on X.

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QuestionLet (X1, d1) and (X2, d2) be two metric space: .

Define d’[(x1,x2),(y1,y2)] : z?=1 d1(X1,)/1). dr 8 metric On X1-X X2.

SolutionHere function d’: X1 >< X2 ——> R is defined as

Xd'[(x1»x2)» ()’1')’2)] = Z?=1 di(xi-Vi)

= d1(x1-X1) '*' d2(1(2»y2) X

SinC€ d1(X1,)/1) + d2(X2,)/2) Z O

d1(x1'J’1)Z 0- 'd2(x2|)’2) Z 0

-' d1, d2 are metrics on X1 and X2 respectively.

.d’((X11x2)| (Y1-)’2)) Z 0

(M2) Let d'((x1,x2), (}’1')’2)) = O => d1(X1-IV1)‘+ d2(x2»)’2) = 0

' ' => d1(.x1-Y1)-‘-" 0' d2(x2-Y2) = O ,

=> X1 = V11 x2 = Y2

. A

Xl

Z’ (X1-X21) = Wzlyz)

Let (951,952) = (ypyz) 3 2x1 = Y1» X2 = Y2 1

X

=> d1(x1-5;) = 0 d2(x2I)’2) = 0 1. _

('1 d1, d2 are metrics on X1& X2 respectively)

=> d1(x1')’1)”+ d)_(xz-3'2) = O ” 1

:>’ d'((X1Ix2:',(}r/3/2)) = O

Thus d'((x1'x2),(Y1J’2)) : 0 <3 (xzwxz) = W1.)/2)

(M3) since d'(<x1.x2>, <y1.y2>) = d1<x1,y1> -+1» d2<x2.y2>

- = d1()’1,x1)+ d2(IV21x2) '

X- V

.=d'(()’1»}’2):':x1-X2)) '

X- ' . d ((y1'y2)'(Z1/Z2))'=.d1()’1'Z1)+d2()@z-Z2)

l 7 » d'((x1:x2)»(Z1'Z2));-*7 d1(x1:Z1) + d2(Xz-Z2) X

»

NQW d'((X1'x2)| (Y1'Y2)) + d'(()’1-Y2); (3112)) ‘= d1(.x1')’1) + d2(x;:)’2)

X

' "1' d] KY1! Z1) + d2()’2' Z2)

““ d1(x1')’1)+d1()’1»Z1)

+d2§x2'J’2) + d2()’21-Z2) A

I 23: d1(x1,z1) + d2 (x2,z2_)

d1, d2 are mefcs on X1 & X2 resp._

d1(x1'Y1) '+ d;()’11Z1) 2 d1(x1'Z1)

I ' 8* d2(x2»)’2) 4‘ dl()’2»Z2) Z d2(x2»Z2) L

' ==“ d’<(x1»x2), (21:32))d’ is a metric on X1 >< X2.

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V Let (X1, d1) and (X2, dz) be iyve inetric spaces.

Let d”[(x1'x2)-()’1»)’2)] = ma-X d1(x1,)'1),d2(x2,)’2) -

Is d ametric0nX1><X2. ‘

' Solution -.

Here function d": X1 >< X2 -—> R is defined as

(M1) since max<d1<x1i_y1> + d2<x2.y2> 2 0)

d1(x1»)’1) Z O» d2(x2»3’2) Z 0. ' i

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d-”[(X1»x2)»(3’1')’2)] 5 max (d1.(X1';V1)1d2(-‘52»)’2))._.----""'G"’~_ ~. ‘

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.‘ d1} dz. are metrics on X1 and X2 respective1§i r ;_~_ ~...-»»--""""'”

Vd”((x1'x2),()’1')’2)) ?- O. l ’ Y

'

‘ (M2) ii Let d~”(4(x1-X2)» (~)’1»V2)) = O :4’ Hwx. ,(€i1(x1-v_}’1)'d2(x2')/2)) : O I

i ' . ‘ i i A :7 d1(x1-.‘/1): O» d2(x2»)’2) = O 31'

:9 X1 : }’1/ X2 = 3'2 i‘-W

1

( d1, dz are metrics on X1,X2 respectively) .

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L/Qt (351,952) = ()’1')’2) =9 x1 = Y1» X2 = Y2 ‘

’ ' => d1(x1')’i) = O» d2(X2»f/2) = O

= max (d1(x1»)"1)»d2(-X2-)’2)) = 0 Q »

VQ => ‘d”((.x1'X2):()’1»y2))= O

Y ’ >_d”('(x1_'x2)'()’i-3/2)) = O <=> (X1-X2) = ()’1')’2)\

(M3) S3159 'd”[(x1|x2)' (y1'.V2)] = max ('d1(x1i)’1), d2(Xz,}’z))1

1

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'max (d1(yi1x'1)»dz.(}'2»x2)) ' '

( £11,612 are metrics on X1, X2 respectively) Y

- (M4) ‘Let.d"[(><i_.xz).(>'1_.yz)]=max(¢i1(»r1,>/1),d2(>r2,)/2)) =d1<X1'..v1) (Say -

‘ d”[(Y'1-)’2)»(Z1,Z2)] = max(d1(Y.i1Z1)- d?v(§":i-Z2)) = d1(.y1»Z1) (5aY)

dII[(x1lx2)| (Z1,Z‘z)] V: max. (\d1£x§vZ]_): d2(X2'VZ2)) ,2 d1_(x1i_Z1) (5ay§ 5}

Since d1 is a metric on X]. . -

\ V‘i1i(x1')’1) ‘IL’ d-1()’1I Z1) Z d1(Xi=?51) H

- 1" ., r .

' : d”ir(x1»x2)I(LV1»Y2)] '1’ Z d-"ii-"’¢'1'-‘$2’-=

J>7\.

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(We get the same result in the remaining cases.)

d” i§ a metric on X1 >< X-,_. .

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d1, dz are metrics on X1,AX2 respectively) - ‘

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Let (X, d) be a metric space and let d’: X ><)( —~> R be given byd(x1 X2) , - .d'(x1,x2) = Prove that d 1s m»e_tr;|<‘ on X.

Solution ‘

Here function d’: X >< X —> R be defined byI d(x1'x2)

'(M) Since —Ei>O X

1 ' 1+d(x1,x2) '-

d(x1»x2) 2 O

" d is a metric onXd (xvxz) Z 0 *

(M2)‘Letd’(x,x)=O=> ii‘-i=0 g 1

1 2 1+d(X1,X2)

' => d(x1,x2) = 0

=> x1 =x2 ('-' a!isI\metTiconX.) /u

. Let ‘x1 f-' x2 => d(x1,x2) = 0 disametric onX.)

UU

C)

I . 1+d(x1,x2) -_

dl x_1»x2 =Thus d (x1,x2) = O <=> xi = X2

. . , d(x,x) ,. 1- >¢¢-(M3) Smce d(x1,x2)-;:d—(;—1j? _-_ 6% I_\) )1?!/4&0

m d(x2,X1)

1*'d(x2.x1)

= d'(x2'X1)

1 d’(X1,X2) = '1+d(X1.X2)

~/ * d(X2'X3)d 1 1+d(X2,X3)

>d'(x1,x3) =ii1+Cl(X1,X3)

/ / _4 d(x1/X2) >A):(z,X3)NQW d (951,942) + Ci (752,963) — 1-‘m+d(x1|x2) + +(-———x2'x3) -

> d(x1-x2)_ + d(Xz,X3)" 1 + d(x,,x2) + cl(x2.'>(‘) 1 + d(x1,x2) +,d(x2,x3)

d(x1,x2) + d(;§'2,x3_1V _

‘ 1 + d(x1,x2) + @:(x2,x,)

*5 d’(x x )+ d’(x ,x ‘) > l(iZ_ .d(X1,>(z)_+ d(1'¢fj) 2 d(X1-X3)1 2 2 3 -'

* 1+d(X1-X3) d 1S3 metnc on X.

= dI(x1'x3)~

I ¥ I ‘Nd is a metric on X.

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LetX = R and d(x1,x2) =A|x1| + lxzl. Show that d is not a metric on R.

Solution »

Letd(x1,x2)=O => |x1|+’|x2|=0 ,. M

=> lx1l: 0, lX2l =

=> x1.= 0, x2=

X 1

LP-'t'X1‘=Xz =>- lX1l =| 2|

i=>

"=> |x1| + |x2| = IXZI + lxzl (Adding Ixzl both sides)

d(X1,Xz) = 2lx2l » -

=> d(x1,x2);= O if Ixzl = O

1 e d(x1, x2) is not always rero

d is not a metric on X;

Question ‘ t .

' Let X = R’ and d(x1,x2) = max (|x1|,|x2|). Show that d is not a

metric on R.

' Solution, V

1' Let d(x1,x2) = O V=> max (|x1|,|x2|) = O

*><I2)

\

=> x =04 x =

'=> x1

, " :> X1:

Let X1 '4: X2 =7’ lx1l:lx‘2l ‘F. M t

= 0, X2 = mm» '»..-»q.,,,_,,___~_______

I |...-.~-..»._...-_.,...._.-__

=> max (\x1l,{x2|) = lxzl “*-‘~'--~"__n___\M‘_W'%KP "

0 _ 1

=> d(x1»X2) = O if lxzl :i

A i.e. d(x1, x2) is not always zero. ll

T‘ius d is not a metric on X.

Suestion ~

Let (X, d) be a metric space and let d :)i >< X -—> R be given by1* ‘K-1'1 12) 1;d"(x x2) = —————'——. Prove that d is e ic on X.

1' 1+ d(x1,x2}

SolutionLet d”(x,x)=O => ~—-———-’———-

l" d(X1X2)

1 2 1 + d(x1,x2)

=> 1- a'(x1,x2) = O .

ml

3 d(x'l|X2.) : \

x2 1

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-cl is a metric onX and d(x],x2) = 0 <:> xi =

Thus d”(x1,x2) = 0 vb x1 = x2

Thus d” is not a metric on X.

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OPEN SPHEREOpen sphere j

with ce

. as S,(xP) == {xix E X. d(x,X.) <-T}

1

1 \ -I

Let (X, cl) be a metric space. Let xo G X and r > 0, then open sphere

ntre at x,, and radius equal to 1" is denoted by Sr (xo) and is defined

A Note .

(i) Let X = {x°,x1,x2,x3,x4,x5,x6,x7,x8} and r > 0

’ Sr(x°) =?

Then by definition of open sphere

(ii) 5r(X.,) § X(iii) A 5r(x<,) ¢ <15 ‘ -

X:2

0

(x ) ._T ° I-= X3,»-.--1...

l 7 ts; —l l 1% A X0 ~; V

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H‘ L " ill X

Sr(x->) = l Xe» x1Ix6 }- \

f ' Here we shall study the open spheres of thetollowing shapes.

(a) Open interval (b) Open disc

~ The shape of an open sphere depends upon the metric space (X, d)xample ' '

\

S_()‘lution ~

(C) Open ball

Let R be the metric space. Let xo =' 1, r == -éfind S1(1).Z

V Here metric space is (R, d), where metric; dz R >< R p——> R is

defined as d(x1,x2) = Ixl — xzl

L VVe know thatIl S,.(x.,) = {xlx E X, d(x,xu) ‘<

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Q Note 4 lI An open sphere in a usual metric space R is always an "open interval”.

Example ' .

1S

' Let the metric space be R2 and let Po 5 (a, b) and r = 1. Find S,(P,,).

SolutionHere metric space is (R, d), where metric dz R >< R —+ R is V

dened as dl(X1'Y1)» (x2»)’2)l = \/(X1 "‘ x2)2 + (Y1 " )’2)2

We know that ..

Sr(Po):{P|PEX» d<P,P@><r) ;1<_5

Put X=R2, P°=(a,b), P=(x,y), r=1

a$1<a,b> = {<><,y>|<x.y> E R2, d(<x[y>. (<1, b>) < 1}

=.{(><.y)|(x,>_') E R2, \/(x - 601+ o - by < 1} r

= { (X,y)l(X,y) E R2, -(X ~a)2 + (Y " blz .< 1

This is an open disc with centre at (a; b)‘ and radius 1.

A . ’ ' » '

M‘ “

4‘I

><

N .

\$\ ‘

2

Nte l - _/ l l

. An open sphere in a usual metric space R2 is always an “open disc".

, ExampleLet the metrieglspace be R2 and d1 be the metric on R2 defined by

d1(P1,P2)=lx1 "»x2| + |)’1 " )’2l-

SolutionLet P, = (0,0) and r = Find s,(x,,). R

Here metric space is (R2-,d1), where metric d1: R2 >< R2 —> R is defined.

as d1[(x1,J’1)' = lx1~ " xzl +. lxl "vX2l -

. ' ' We know that ‘

. S,(P,,) ={P|P eX, d(P,~I-2,) <r} ‘ ' ‘

Pat X =‘R2, PO = (0,0), ‘P = (x,y), r = .

5i(O'O)={(x'y)|(x'y)ER2' d1o(<x,>->.<o.<>>) <—

\ _.

» ={<x.y>|<><.>»>eR2. -"T:l<1} R

' i ”\/2 .

This is an open square with x-intercepts

r'-~\r-'-\r-*—\

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Let (X, do) be a discrete metric space. Lei: x,Ye X and r > O

- Find Sr(x,,), when (ii) rél (ii) r >1Solution

Here metric space is (X, do), where do: X >< X —> R is defined by

< O i x =x ~

d°(x1'x2):l1l fjljxiaxjWe know that .

S,(x°) = {xlx e X, d(x,x°) < 1"} — *>—-—-— (1)

be .

Whenr<1

If x at xo then from equation (1) we g it )1 < 1" (False) 1

If x = x,‘ then from equation (1) we g -t Q < r (True)

Thus Sr(x,) = {xlx e X, x = xo]-=-1.x,,} - W

VVhenr>1

'1‘kill: \

Note

:1“ -

(1)II

If x i x, then from equation (1) we gi 1: 1 < 1" (True)

‘ If x = x., then from equation (1) we g 'L' 0 < r (True)

Thus ST(x,,‘) = {x|x e X, x = xa or Xa x,,}

={x|xeX, x=x°]U{_ l:\xeX, xixo}={xo}UX—{x,,}

From above example we conclude that:

1 An open sphere with radius less than or e ual to 1 in a discrete

metric space is always singleton.

l itl adius reater than ' in a discrete metric space(ii) An open spqere w .1 1' g

is always the full space X.

f .

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QuestionLet C be the set of all complex numbers and let dz C >< C —> R bei"“"' ' 1;;

defined by d(z1,z2) =_ lzl — zgl. Find Sr(x.,) when xo = 1, r = 0.01 -»

‘JSolution

The given metric space is (C, d), where dz C >< C ,—-> R be defined by

17

d(Z1'Z2) =4lZ1 _' Zzl

‘ ~Now S,(x.,) = [x|x e X, d(x,x,,)h < 1"}

PutX=C, x,,=1 r=0.01

1

9'_, 1

T if ‘ 1: T

.

‘ ,,, ,3 I/.-' l

$@.o"1(1)={><\xEC.,d(x,1)<0-01}1 ,1? 14;

= {xlx E c, Ix - 1| < 0.01} --————— (1)/u.-//4 00!

Since~xeC x=a+ib=> x-—'1=a+ib—1

=> x—1=(a—'-l)+ib ‘x4,.ol} .

I.

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Question

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at!-'l-rv-*>!}""7'°\\\f lT

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(1) =>s,,_,,1(1) = { (Cl + ib)|(a +-'11») E c, ./(a - 192 + b2 < 0701} ~

z , ‘I

={(a+ib)|(a+ib)eC,(a—f1)2+(b—-O)2 <(.01)'} -

Let d be a metric on X and let d’:X >< X —> R be giveniby . if

..d’(x1,x2) = min (1, d(x1,x2)). Describe S,.(x°).

Solution .. .

4 -

Here given metric space is (X, d’), where d':X >< ——> R be given by I T'5

Now

Question

d'(x1,x2) = min (1,d(x1,x2)

. = {xlx e X, min (1_.d(x1,x2)) <-2"}

This is the required open sphere.

. \

l ll;

V‘Yr!

. ‘ H

5r(x<>) = {XIX E X, d’(X"?») < Tl '

' Let (X, d) be a metric space and let d’: X >< X '—> R be given byd(X1,Yz)

V T d'(_x1,xz) =;—+-;1—(;';;5. Describe S,(x?).

Solution

r

f Here given metric space is (X, d’), where d’:X >< X —> R be given by .

\ Now

1 - ~- ~- -~ 4- e --’_ -- i »+-_-__-_ .___:1,.._,_._

'\ ('i(x1»x'_’) .

'1

-d’ rc x -——————-+

1' 2) 1+d-(x1,x"2‘J

ST(x,,) = {x\x E X, d’(x,x,,) < r}ll Y

I

: ixlxie I ' 1 + d(x1,xz)

This is the required open sphere.

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. Let x1, x2 _be any two distinct points of a nr rtric space X. Prove that “‘~\

. / there exist t1/vo open spheres Sr1(x1) and 5 Z(xZ) in X such that

a sT1<x1>n s,.2<x2> ¢ él 7

eet' Let S,1(x1) and Sr2(x2)

- be two open spheres with centers _A - _

\'<

xi and x2 and radii r1 and ‘r2 ‘

A respectively.»

' I I Ii; I’ -l I‘ if C

‘ Leta d(x1,x2) = 13 + r2 ii'~ X; tin i 4

L 2, IWe are to prove that < Y (x1)ST‘ (xi) n srzcm = <1» 1"”

l 1

‘A "' J We shall prove it by . ~ {

contradiction method. -

I . ‘ SupposeA 5r1(x1) fl Sr2(x2) ¢ qb » - 5 .

Let x e Sr1(x1) Fl Sr2(x2) -

=> ix E ST1(x1) and x E Sr2(x2) i . ' /X=>p d(x,x1) < r] and ci(x,x2) < r2- (1§_X,d)

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=> T1 + T2 < T1 + T2 [BY (1)l ' (U2; PW‘/3 $;;0(r)/7.5§§'_xa»)

lt is impossible. i nu. >tf_-_,S_’.z"(7(l) Sy.(>(_-)

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L‘ PEN SET‘ ‘ ~

Let (X, cl) be a metric space. Let U Q X. The U is called an open set, if t

for each x E U ‘El 1" > 0, such that S,(x) E U. " ..

i.e. U is called an open set, if each point of U is the centre of some open

sphere, which is contained in U. * ~'

¢ »-

4

1

X L. xi ,i/ ~

H-I t Q

f" . -.~ .

1

1 *~ 0

.1

.. W E7'*"_x

U is an open set. U is not an open set.

// Example V

..,

V»

Let R be a usual Inetric space (The ordinary real number line) and

let U=]0, 1[, then show that U is open. ‘-, .

S0luti_o_11 '-

I L

Here metric space is (R, d), where dz _R >< R —> R is given by

d(x1_,x2)=|x1 ‘“ xzll ,

-, Let x, e U/i Let _r> 0 »

"Then ST(x,,) = {xlx e R, d(x,x,,) < 1"} .

ii‘: {xlx e R,“ lx — xol < r}“ A ~

‘

={x\xER, x—x,,<r, x—x,,">—1"}

={x[xeR, x<x,,+r-, x>x.,-=r}

_={x|xeR,x,,—r<_x<x,,+'"r} t

"- W

. _. ........-____,,. 3

i=]x,,—-r, x.,+r[

. l/W.» can find a value of 1" for which Sr (xo) =]x,, — r, xoi + ’r[ Q U=]O,1[ *

Thus U=]O, 1[ is an open set. l_

Note

' Then sU‘_O01(0..99) =]0.99 - 0001,09? + 0.001[=10.981,o.ss1[‘; ]0,1{

Ix

2

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M

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l<

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=:i;,~";::;"_*§._____T.__Ah,

li.l

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~ 1

In the above example if we take X0 = 0.99. Let r = 0.00] 'li

ll

1

kk--,_.=+'.<—-—*

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\F.

ill/1\;~

‘K 4) Let U= {(x,y)|(x,y) e R2,x2+y2'»< 1]-.E1§l1owthat U is an open set

ExampleLet R2 be a usual metric space (The ordinary‘real plane)

Solution

»:d)b& Q Here metric space is (R2, d), where dz R'2>< If-—)R is given iby

‘ I ,2 ’K, _(/451., d[(x1'y1) (X2 )’2)l = \/(Y1 _ X2)? +'(,;V1 _ 3/ )2

3,3 /W1/e A! Letd(0,R,)=/1Let r_= 1 — /1, then r > O

I F

Y\,,_

3:5.&M~*‘ta V__

4!‘

.1.it[W

s»

‘ 1:‘

4:(‘

, t

= t‘ 1

::'§*F‘

l‘.uxv;

J»-1:.\,.51

/as“ We shall prove that S,.(P,,) E U

. - ' - I 2

‘ W)l ///5% 7' Qt ~ E U-

/ /"W/ 3&1‘ gs » \ .- 1

= ' " “" “' '?l§‘l~5'i'~4"'f§'.:

I\ , I

» 4‘ft - R3“!,,..~,.§4 1;».

"vi \ ehL 1 ».»*"I

, /-/'*5“?

~1

l

auctw *’/”"p Let P e Sr(P°) => d(P, H) < 11 Q 0(_0.0)Ec¢)7l/Q’ 0 )‘ lr;;;;~;~‘t:'

1 u; — 1'——,/i+A>d(P,0) r=1—'_'P,O)<1

‘ -pre/>24" - PeU

I)1

ii‘, z. /5 S'nce P e.S‘,(P,,) => P E U.L/ .-.1 gr (11) g U 1‘-Q”-.3“

Hence U is an 0 en set. '"’41:11" 1 PLxam e

Vi?) Since (R2, d) is a men-ic space, \r‘ '- d(P. -'1) + d(P°-; 0) 2 d(P. 0) ' 7 /R2

» Fis ‘C)) ' -' " ' ;-‘4,, /é {I " => r_+ 21 > d(P, 0) r~>¢Z(V,/4) .e:%._.1;;..!';,-__;.;/

:‘-.

,\I -.

V

/1_

'1

- Q’ Let R be a usual metric space (The ordinarj real number line) and"1 let U: {xix E R, O S x < 1}, then show tliiatUis<€;en.,-M" m0.@oIuti0n

»

-“ '7. Here metric space is (R, d), where dz R >< R —> R is given By

14v-

//'

/@4);¢/as/)d<x1,x2>=»»~-xzi , ,.<>s<0//)4 Y5 U={x|xe_R,<0§x<l%} /[0/6) /6=[0,1[

u ' "E )é(-I’/72Let.1c°='0EU, Letbr>0

lhenS(O)={x|xeR,d(x,0)<r} A/_/_/_p¢;/p 1} '

AW,g,“~l\‘!‘ 7 T ' .hilt/0/,)éi ={x|xER, |x—O|< "

g} = {xlx e R, Ix! <2 rjH .ip ={x]xER,x<", 1>~—}

‘={x|xER, —?”,<~::<'I'r}l " =]—r +r['i V We can find a value of 1" for which Sr (O) .=j ~-- r +'r'[T.£ U=_[0 1[

M95;,H Thus U=[O, 1[ is not anopen set.

l (/3) 'L

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Theorem ' T

’

Every non“-‘empty subset of a discrete metric .space is open. '

Proof *

Let (X, d,) be a discrete metric space.

I Let U <;'X'such that U ¢ <1» V"

We shall prove that U is anopen set. l

»

|T l

-Letx<,EU. - ‘

3

ii / ’ -1

LetO<r<1 ‘

1 Then ST(x.,) '=. {xpc E X, d(x,x°) < r}.

_,:{x°}‘_

The open sphere in a discrete metric space, whose radius is less

T Since 5,05,) 4

Z=> Uisanopenset. .

V

VExample‘

- Let R be a usual metric space (The ordinary real number line) and

let U= { 0 }, then show that U isZg{pen; V.

Solution " ..

.

Here metric space is (R, d), where dz R >< R -> R isgiven by ' l'

Here U= {O}

“Then Sr(0)'= {xlx E R-, d(x_,0) .< r} 5i

={x|xeR, |x|<r}‘ — ~ kl

={x|xeR,x<r, x>——r}

={x|xeR, ~—r<x<r}=]—r, +r[

We can find a value of r for which ST(O) =] — 1", >+r[ Q U= {O } V

» Thus U: { 0 } is not an open set. ~- '

'

\

- T '--—-—"-———-_—————'I“‘>

O

T.

£,,.. Vc W ’ T ,1’?-i-Y~1» §l'.1I'_-“ " ; * I

4

d(x1-x2)=lx1- XzlT

- Letx°=0eU, Let"r>Ol

5

Y

ItV

.

_ ={X\XER,lX—O\<T} 1

l

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Union of any collection { Ua A: a e I } oft) pen sets is open.P ()

ii;Y

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...‘\.,t.11.!Mr}

4VA ‘LMtg‘I1

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rt}<:

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X

n “

TheoremLet (X, d) be a metric space, then

ii Intersection of finite number of open. sets is open.

(iii) The VVhole space X and the empty set (pare both open.

Proof '

it Let { Ua 1 a E I } be any collection of open <3'ets in (X, d).

We are to prove that, Q3 Ua. is an open set. .

Let x E 0,3 Ua

Then ;g_ e Ua for some or e I- : .

Since each U0, is an open set therefore there e><ist~ r > 0

1 5‘ Such that ST(x) Q Ua for some oz e I _

g:‘ 5r(x) Q 0ceUI Ua

‘=> 'aeU, Ua is an open set.

(ii) Let { Ua = cr = 1,2, ..... .. n } be finite collecticn of open sets in (X, cl).

. . n ,

We are to prove that O U C, is an open set. / '

a =n E

Let xe NU“c:=1

= xeUa va=t1 .... -n L’it 7

\ \,

v-Since each Ua is an open set therefore theie exist r > O ~

_ Such that ‘ Sra(x) E U0, V a = 1,2, .... ..n

Let“ 1" = min { r1,r2,r3, ....urn}

_\. -

iii“? L => S,(x) E 1,1,1 ‘v' cr = 1,2, .... ..n

tr HW’? => O Ua IS an open set.

M a\It I

(iii) To show that empty set (1) is an open set, we have to show that each

point in ¢> is the centre of some open sphere which is contained in qb.

But since there is no oint in , the iC()f1<.§i,ifi(\n is automaticallP Y

(.5 ;.~ satisfied.

Hence c/J is an open set. .

Since every open sphere centered at a point of X is contained in X.

X is an open set.

1],.‘ It‘ ik‘ ~ ~ - . . ...__,.____,__,Mr;...‘..1~*.“ -~ \ ,' . _ ‘ " :"7'.:_.:."_.,_——Tv,?_.-¢-1-;y_~_~_'_ -

11

ll

\ ' LL ii 4

“pa.

An open sphere in a metric space (X, d) is an open set. ‘."~.. ‘

Maltt€%.%3I.0rg

Merg'ngManandmaths\‘S

s7’ 23 '

Proof ' I L

-

Let 5,. (xo) be an open sphere in (X, d).

- Let x’ E S,(x.,) => d(x’,x.,) < 1". ~

.. [.i"1:*. i 3!?"

'

<

P ,-;‘z~‘1i$i.t*>;~.- ,7

, I _-_

- . Let d(»< -X») — /1 "

I I i -V P

Let T = 1" — )l, then r > 0

A i We shall prove that 5'r,(x’) Q S,.(x,,)V; I *

~

Since (X d) 1S a metric space, Sr’ (x ) 7 "‘~ >_ . \..

‘"¢}"‘s.~ ’,..;?i§1;e;g-,1,~,3g -‘ -. . V ;

. - Let x E 5./<x'> = d<><.x'> < T’ T "~ . rt<-~~w1- :..¢‘\' I ‘ _ . ' -

- I I #3;M.,;1~,.‘ xélvptw... .

r ,) A

#> r’ + A > d(x,x,,) ,4, s A C»1_,o¢) 5,9:

i=> r—A‘+,1>‘d(x,x.) hr’:-r—'/1 "~

='> d(x,x.,) < r

Since x e Sr/(x’) V=> x e ST x »

' Sr’(x ) Q Sr(xo)

Thus ST (xo) is an open set.

Hence open sphere in a metric space is an open set.

heorem ’ ~

' ' '

A subset U of a metric space X is open if and only if U is union of

open spheres.'

Proof‘

Let (X, d) be a metric space. Let U Q X. We have to prove that

U is an open set <=> U is the union of open spheres.

We suppose that UL is an open set. Since U is open therefore each

point of U is the centre of some open sphere which is contained in U. i

.

Thus U is the union of open spheres.

Conversely suppose that U is the union of open spheres. Thus bis

,1 the union of open sets. Open spheres in metric space are open

sets. ) -»

-.

Since the union of any number of open setsin a metric space is an

open set. Thus U is an open set. '

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.

Let X be a metric space and let {xo} be a singleton subset of X.Then X — {xo} is open. I

Proof »

'

Let x e X — { xo}

' “ ? ,“:.'v2I.»‘¢. ~_.~'-1.

imde»»=¢ --------4D /’ €*%@&$~v

| xvWe shall prove that

Sr Q T '‘ E Sr . "

=> d (X, X) < 7" ______ ... ' V i

From ('1) and (2) we get " ,. ._ e '

=> d(x’,x) at d(x,x,,) ,

‘ => d(x,x,’) at d(x,x,,) ['1 disametric on X. So d(x',x) = d(x,x’)]_

~

1-*1

>4O

\-.,-J

4*;

.»

Q

‘CW.,

1-:

/}»".§.’"+

'155;‘;-.' ,..'

‘.ill'=\.)

s*.»="iI-a.*"'-“I;~‘1"'~;£‘~”~'¢.4Q.-1,,

.'<~1.h!"G1,-».\,,$,-,-J?

,

\ ~

=1 If => x’ at xoHi;

= x V65 {x,]J => x’ e X

% —{_Xo}

fggg ' Since x’ e S,.(;_¢) => x’ e X — {x0} '

; €9 .=54@;X~gn}' “g

‘ F‘ Since every point x of A ~ {x°}1s the centre of some open Spite .

s ”’“ '‘

.1 '* ' ‘Rt

“ti 5“ contained in X — {xo}. ' '

ll X

‘W ‘ Hence X — { xo} 1S an open set.I ' ,i::;i it .¢ Question K)»;

, Can a finite subset of a metric space be op€n? Y

T Solution

We know thatl ‘;

iii; If (X, dc) is a discrete metric space, then ever/v subset of X is open.

Therefore a finite subset of a metric space is open. -

it ,_ i (ii) If (R, d) is a usual metric space then {O} £2" R, is not open.

T » Therefore a finite subset {O} of R is not open.-

Thus in general, we can say that, finite subset of a metric space may

or may not open. ' -

I 2 Metric Topology

The topology determined by a metric iscalled ”metric topology”.‘L, ~ .

7,5,‘:33;.=»,3,3_:i§f_{iy§;§+=|;¢1..=;...,.,~<_.~,.M-+a-..~,..,-...i....¢,,-. .. ..,__.__.._._......~.,...i,...i.i,.__.___.__.._,._,__ ...,-__ ,

§OR

ll I!.

’ Pro f . <

‘

T

3

C41-4J¢ 25

E

Vat/)3»/-c i

The rem .

T is a collection of all open sets in a metric space (X, d), then T is‘.

at pology on X. A -

' \

. A m ic space is a topological space.

' ' Let be the collection of all open sets in a metric space (X, d).

. e are to p ovepthat, T is a topology on X.

__(T1) -Let U0, T Vael -

.=>' aisano enset. Vael=>. a Uaiso en. »

E( nion of ny number of open sets is open.)

- U , .

vl

=:~ aE,Ua ET .

5

' (T2) Let Ua ET V a = 1,2, ..._n I »

*2;

.Q.)

T ' n - _.

‘ I“

i=>" O Ua is an ope set. ., -A

“ l

a=1 -

~

.(1 ( Interse tion of fi ite number of open sets is open.) ‘ .

a

|

l

1

<

1

. n

=> H U e T ( definition of T) -

~

aa=1 V

. .

(T3) Since ¢,X both are open. V

T» q, X e T ( B definition of T) -

.

Y

Thus T is a topolog on X. '..

i.e. (X, T) is a topolo ical space. ’ -~

' This shows that a ” etric space” is a "topological space” whose

topology is "metric top logy”. ' -

TheoremEvery non-empty se can be given imetric topology.

Proof ~ '

We know that \

(i) Every non-empty set can b given a metric nd can be converted into

metric space. .

(ii) Every "metric space" is a "to ological spac ” whose topology is ap

"metric topology".‘

Thus from (i) and (ii) we conclu e that every on-empty set can be

given a metric topology. \ .

.‘

‘. ‘

-.»--

v

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ar.1

ti.2

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:2 => xJ

i

-Now Sr(x’) = {xlx e R, d(x,x’) < r }

ti ' ={mxeR»x~xw<r1I

i

l

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‘ii ii:

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, .If K _;-; 26Ni’

CLOSED ig 1)

Closed Set

Let (X, d) be a metric space. Let F Q X.

Then F is closed ¢=> F’ = X — F is open

‘ <1‘-~ Example Ca W»,-ll‘ Ewxgls/"\\

’§ 1‘ ‘ \V

g\$°\§\

<2,&Q%‘\<§.F

,7 ' LetX = R be the metric space and let A ;='Ca, blwhere a,b e R, &a < b. Show that A_ is closed set.

Solution " ' .

The given metric space is (R, d), where at: R;<R —> R is given by

d(xl'x2):ix1 —x2iSince A = [a, b]

I OA =R—[_a,

=]"°°,a[U]b-°°[

‘#5

F

-14

ii

R

Z La i "00 ‘ _ ' ~_-> w

In order to prove A is closed; we will have-13> prove that A’ is openf‘ll .

"1" Let x’eA’ => x’eLaUR,, ‘ -"' I

V => x’eLa or 2g’eRb '

Case-I If x’ e La then x’ < d

ii! ii; F Let d(x’,a) = r

1| I -a=—r x’<aH‘M = M+T=a ----------------------------w4n

= {xlx e R, x’_-_r <_xp<_I’x'-l~’r}V .

=]x'fr, x'+r[i =] x’ —'_i', a’[ [By 1

/

O 4 l‘ ~x-/ ~ ~ —-—-- -—

/ Thus Sr(x’) =] x’ —'r, a 14 E La Q La U Rb =2: A’

5 tx’) Q

Hence in this case A is open.

i L,s'_ <5/,~i ={x|xER,x-x’<r,x—;((>—r} ,

Ni?!‘ O\ .

={x|xER,x<x’+r,_x_>7('_—1f} i

iii‘! A/O

e K," (1)1 <1

/\_/~A /'/-73‘

> 45$ ‘Q9

F

3

v

i

I

'‘ >27

' ' Case-II If x’ e Rb then x’ > b

‘V ' Let d(x’,b) =1"

i‘ '=> IX’--b|=T' => x’-b_=r x’>b

. I Now S,(x’)'= {xlx E Rt, d(x,x’) < 1"} ‘

I ‘I ={'x|xeR,lx—x'|<r} A

A" ={x|xeR,x—-x’<r,x—x'>—-1"] .

‘ »

={’x|xe_R,x<x’+r,x>x'—r} ' " W"'

I

={x|xeR, x’—r <x<x’+r~}

=]'b,»X'+T[ [By(2)1 \

Q

B"

1

\u

iI

A

i

/fi\\I

I \If I T \

, , ,___._..__._._>

4 L \\ x ' I1

‘ !r- FM»-

/Thus S, (xi) 4;:-1] b, x’ + 1" [ Q Rb Q La U R? ,= A’ """‘j"“‘""""--*~ _ .

i1.8, s,(x');A'._ _ I " " -' "

I

0 I ' n

J

*

Hence III this case A’ 1S also open. . ., ,

s

'‘

» 7 Since in both the cases A’ is open. Therefore‘ A is closed set. ~-

a. le I

V

‘ ‘ Let Rzcbe the metric space. V'

I Let F =4 (x,y)l(x.y) E R2. (x — av + (y - 11>’ $ 1}.» ' Show that F is closed set. “ " .

Solution ' II A

I

" Here given metric space is (R2, d) where-dz R2 >< R2 —-> R is .

8iVe1'\ bY d[(x1»Y1)» (_X2»)’2)] = \/(X1 " x2)2 '*f'(Y1 _ )’2)2

?E

Her? F = i(X')’).|(X;Y) E R2, (X - 602 + (y — .b)2 S1

I Thus Fs'={<x,y>1<x.y>ER?. (X-a>2+<y-b>2>1}In order to p_rove thet F is closed, we will show that F’ is ogen.

§ .

Let _P' e F’. Let d(P',-Po) = A "“"'""\

_- .,. .¢ ~i-;._;I._.

-

- -I -;,‘-\=~J»"».\. ~: ».~,-- ~- i

I

F ‘ '- 'I‘

Let r = /1 — 1, clearly 1" > O ' _ ’ I

,

~' ;5;‘\1,»}3§

We shal1_ prove that _§‘T(1_?’) Q F ’_ K I »

\ ~~ ‘ ‘ * ,. ’V::'-:;;:‘:’i“VE

I‘ I I 3 : \ ‘ , *”"' ~‘

Let P E Sr(P) => d(P,P ) < T ;F ;

“J

i a

=]x’—r,x’+_r[v L

w

A

<

.-‘. ~ Y‘ _ _-it __ ._~;..U.:_J:v

I

1

" Since d is a metric on R2 ' EQ11 *3) A"Sfi(Pf)‘.i::?.v‘:‘\

L

d(P’,P) + d(P,P.,),2 d(P’,P°) I R2.,.,<_-.-ex»-1 gg ‘.,~ ?,‘.___._;_v- ,_ ,1»

I s.

\

~—,:- » - -Y»->1-V»-_-—-—=---~~,s==_,G,-,,,---——.__.;_._.._.., I I ..,v_,_..-_, _____.

---~,_— —

V

' ~ ' ——— __;;— ~¢~ — 5 — ——_ _ ~ ~~-V

'\__.i;___-__

U

><_3'

‘I’.

B;

IIIIIIII5IIIIIIIIIIII

/5

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I

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;~.>'&‘-'\\h8?.-l‘‘\\ -e<‘

‘_'-fa( y.._,_V3 _‘~_,3-§%_Y

4IE*‘:'~‘3;ite’' .=;-'1;-1*’

<~;I_;=;‘

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I \ \"\

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lrL

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la‘;

lll

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22.

M , /1 "Z, T + d(p'p°) > A _ /L 7 0((,,l7//F/C,,7¢jl§//Cf an qven Sf Lea

7‘,/ed f /Qavr/A64/h<‘j '

=* => d(P,P.,)>/1-1" Ce)?

=> d(P,R,)> ,1-(,1-1)'=1 ’§”%‘” /L-57“/?’%//_4% =.~ d(P,P;)>1 A

A r./—(‘/1-/) i dorm)’/I

~ lill?’ /I

l SinceP P e F’ 7/‘aka ‘<7/(P2 .

l sr<P“>‘é/F’ /cw» /AM 5/(/1 P )4 w=.~ /Ti’ is anopenset. ad’/4'=> F is closed set. F9” fa; w¢6ef ,/@491 it/d>3’*(¢

Ewpe _t ; /if/2’t_,/_Let R2 be the metnc space. WM 7”’ mg

s_<>11:io_n ‘ age,/i) .4

9* A = ‘Z (x-J’)|(X, y) E R2, X2 + yz S1}b€.a subset of R2. .

IsAac1osedsetinR2? dfe/p) i4(/%/P/)/'d(P/P)

2\ >

I

1

lI

\

1/ x

A+/0Here given metric space is (R2 d) whee ye d- R >< R2 -> R is

given bl’ ‘ dl(x1'}’1),(X2')’z)l = \/(X1 —§c;,_)1*+(y1 — y2)2

Hére A-= { (X.y)l(X.y) E R2. X2 +y2_€ 1} L}; /i’ /4’!' Thus /1’ ='{ (x,y)|(x,y) e R2, x2 -I--y2_ > 1} 5:/[/;,)'é/ll +(J-4)

.In order to prove that A is cl0sed,’we will show that A is open. ‘

l am, P)4 1Let P’ E /1'-' Let d(P', 0) =/1 /”“' V,>.,1 wk): _. N . ‘

I . . .~> 4 >.f$=A.i§».§;;_ Q FLet r = /1 —-1, clearly r > 0 '2

We shall prove that Sr (P’) Q A’ I V ‘A in .:);i3il;f5 Fé gxp )Let P E $,-(P’) => d(P,P’) < r

V Since d is a metric on R2

d<P¢ P> + d<P' @> 2 d<P»'. 0)

J ~F§§/(7),(P') F 1» 01/WA

f.+ d(P,U)>/lp' r

l

= vd(P,O)>/1-1"

. => d(P,0)>/l—(/1—1)=1=> d(P,O) >1

l~ i=> Pe/1’.1 "\'\ '

Since P e.:S,(P'),l => P e A’

5,(R’)'"; A’

is an open set.

i is closed set

1

l

i

I

I

1

I

i

A

v

iii

.

I

t

i1

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i

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29 =_

1.

I p‘

-' - Example ~. ' -

LetR be the real line and let A = {xlx e R, 0 3 x < 1 }, be a subset

“ of R. Show that A is not closed. Q

Solution -'

The given metric space is (R, d), where dz R >< R -—> R is given by

d(x1»x2) = [X1 " X2‘ U

i X= [o,1[

A’=R—A ‘

=1-<><=.<>[u[1.<><>i, -

Note that 1 e A’. We take x,, ="1, and r > 0 '

I \ . -

7--— .-._......-..¢.:..-..»- m.-<w.‘-..-.-m-._-.__._ _.___.

.. 2 . ‘

'>

I .=_<v _.

Put x.,=1, and X= _,.Then S,(x,) 7-’ {xix E X, d(x,x;,) < r} 1 " " A , '_ V’

-

i

R “ A 9 ;|

- at\

'' 1. :,, ‘cw . .

={x|xeR, |x—1|<r}={x|xeR,x—l_<r,x—1>-_-1"}

/ ={x|xeR, x<1+r, x>-1—r}

={x|xeR,1—r<x<1+r} . ,

=]1_"T11.+T[ButST(1)=]1—-r,1+r[$A’ vr>p0’ _s,(1

.. ..) i

Q Thus -A’ is not open. » ( ’, -\/'’ \ 3

‘ => Aisnotclosed. :00 , »‘K J/~

Theorem-

A subset U of a metric space is open if and only if X -— U is closed.

Proof An S“b'<°tF ” ll ’ ’/ ” »” ¢’ »;’(’(’.’.’Y*?i‘l.

1 Let (X, d)be a metric space. We have to prove that V T

i U is open ¢:-> X -— U is closed. '

‘Suppose Uris an open set. /i Then (X — U)’ -= (U’)’ ‘.9 XA4 "’ M 1

’ =U (Openset) t

Since (X — U)’ is an open set. ‘ ’K

X - U is a closed set.

\ t Conversely suppose that X — U is a closeti set.

Then (X — U)’ is an open set. ~

=> (U')’ is an Qpen set. _X —- U = U’

‘ =_-> U is an Open set. ‘R

' /* . ,-t__._;1- . .. I .... .. ..

xi. .

.

.

* ' Here AV-={xlxeR,0$x<1} ,

.

1

51(1) = {XIX E R, d(X»_l) < T} ’-___.__’ ..-__;-.....-}

V

l l

,.

:in'_t::;_:r:'

\

1

R

R

1

l I

1, r

H;

Np WL: (ii),\ I ('1 )

QM;‘,|r.

vi‘

‘i‘Kl

L 1

i

I

I 1

l

.

51'

I ~|§'i

W ‘ii’iii] N’

.j ii

‘ii:

ii!Pr,

H .iHi

k

iiiH1

i‘ '1

_Tr*\f

"‘.—é-§%_‘—i1:-_~;~:i‘¥E:_=;:.-4

i ‘:1 l\-

L ii‘

I3‘-iiill‘;

§i~:

'22K;£1"

‘ll5..

r‘- (

.,.

ll ,

\\

' 30

Theorem

1'iProof

<1)

ll. .

Let X be a metric space. L

Intersection of any collection { F0, = a E 1} of closed sets i

Union of finite collection { F1,_iF2, .... .. , Fa} of closed set is

X ‘and ¢ are closed.

Let {Fa = 0c E I } be any collection of closgdisets in (X, d). <

X'-.

s closed.closed.

“/\,.@¢g“

\_JI§ 6

Then Fa’ is open. V oz e I 1:

u ,1:7 ael Fa

~ n ’- .. u _ n -’=> (as, Fa) 1S open. - ael Fa - (e, Fa)

\

LH

ael Fa

is open. Union of any ninnb|:»r of open sets is open) {fa-’_

9-RD

H

isclosed. £4 :En Fina‘ " ‘ ’

ii) Let {Fa = gr = 1,2, , n} be any finite colleetion of closed sets in (X, cl).o

Then Fa’ is open. V a = 1,2,...,n

n ,:> O F

:c=1O, lS open. Intersection of fini number oi open s».\';s is open)

n ' n n ~’

= < u 5% mogm. -= n @'=< u Q)ii‘ ct-=1. a ‘1 a 1

ii”=> U Fa is closed. a

’a - 1

(iii) Since ¢’. = X — <1) = X which is open.

=-> gb is closed.

And X = X -e X = <1) which is open.l

W Ill!‘

=> X is closed. '

Ql1€StlO1‘1

Is N close

Solution

dinR?

, I

Here N = {1,2,3, .... .. }

=R+N.=] —O<>,1[U]1,2[ U ]2,3[U..,..

= Union of open intervals in R

= U11iOn Of open S€fS /‘in open intervalt in R is an open set)

= OPGP. set Union of any number ol open sets is an open set )

iii ii Since N’ is an open set. --

-_~:» N is aclosed set.

Ihpms ;'\~¢+~.. _ - -

I»|l-l \

~-

. . '2 ,7:6 1

.4.‘

§

I

it‘.1A

H1\ti

H'1>

grew»/*’“*~*’:“*¢"*,1:”~’</'>*-<-wt-»i~~.4<><,¢» M»; A

-1”‘? KO M ‘r-7»;-'~»*‘<z)///" =97/4 m Z’: ¢‘5’»!~/Q-’2<.¢ *

~

~ 9 ' /A >

‘t

l

1

Mllt€iIY,0I‘2

Me"9"QManandmaths

-

Limit PointLet (X, dy) be a metric space.”'Let A Q X and xo e X. Then xo is

/1-

/f LIMI'€‘gOh€;/I‘

Y?

~ a

oft»; .7/pug»/6

called limit point of A if each open sphere centered at xo contains at least

one point of A different from xo.

s,1<x.,> .S2(x°) it Sr1(x) 5r,(XJ) Sr1(x<>) Sr2(x»)~ - V

1

i’ l

l

4

____t__,___c___~._it._-,

i

1

A

,_

>’ I

Q

_‘Q I-_\i

I llI’-__\I,’,\

'1

\\

\.~._

‘ ~_’

U_____,____/

d——\-

’_;;>///I\

I_\ I\ \U

\5I\__/

\__Q1~__,’

.

1

l¢~ wet . .W? Va ~

:~<1‘ - - l

; . 1' ‘ ~‘

\

1‘ 1 \ \

. I I \ i\‘ A» )

'\ -x/' .4 ~"‘\__/I 1 , / -—~f\

\——’ I-Q\_ ’/ \

L’ X - I I - ' \\ *~v '1.

__,,, X \i____,, X F

s___ \___¢

____ ~_________

(Fig — 1) Sr3(xo) A (Fig — 2) ST3(-xv) (Fig ~ 3) s,,<xr>A

In Fig—1, x, is a limit point of A.

In’Fig-2, x, is also a limit point of A.

£3?i TIn Fig—?>, x,.is not a limit point of A.

\/ 4 Let (X, d)tbe a discrete metric space. Let

7 point.Proof

Consider the discrete metric space (X, do).

Here do: X >< X —+ R is defined by

Theorem '

A 1; Then A has-no limit\ ..

:

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IJ . ‘ .

We have to prove that, A E X has no limit poi'nt.-~

. We shall prove it by contradiction method.

Suppose x, e X such that xo is a limit point oi 21..

Let 0 < r < 1 then S,(x°) = {X0} <1)

In a discrete metric space the open sphere with radiusless than 1 is

always singleton. 1

Here (1) shows that S, (xo) contains no point of A different from x

Thus xo is not a limit point of A.

Hence A has no limit point.

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32 1

Let R be the metric space. Let A = {xix E R‘ x =:11,n e N} be a

subset of R. Show that "0" is a limit ptlint of A.

SolutionHere metric space is (R, d), where dz R >< R ——> R be defined by

d(x1»x2) = lxl " xzlll;1 1 1 1

' Here A=<,l1,—,—,-, .... ..

Then S,(O) = {x x e R, d(x,_O) < 1"}; r > O

'_={xxeR, |x|'<r}=[xxeR,x<r, x>—r}={xxeR, —r<x<r}=]—r,+r[

" Clearly for every r > 0, Sr (O) =] — 1", +:r" [cnntains a point of A

different from "O". .

YQ stion

011:1 I, .

iii!“ Thus "0" is the limit point of A. A

i ll‘ , ue V

1 ";,‘R i Let R be the mei1"ic space. Let A = {x|x e R,x == 1 orlx = 1 + — ',n e

Tl

be a subset of R. Show that 1" is a limit poirt of A. ,

5_0_111_!Yi2I1 . A

He metric space is (R, d), where dz Rx R -—> R be defined by

d(x1»x2) = lxl " xzl

Here A={x|xeR,x=1orx=1+%,ne1\/EL

={x|xeR,x=1]U{x|xeR,x=1+1/n,neN}3 4- y.

1 ={1} U .

A 3 4

- ll‘-2"; lNow ST(1) = {xlx e R, d(x, 1) < r}

-={x|xER,|x—1|<r} 1

={x|x€R, x-1<r, x—1>-~r}={x|xeR,x<1+r, x>1—y}={x|xeR,1—r<x<1+1"}=]1——r,1+r[ 1

Ava'lableat‘

www.mathc'ty.0f9

Clearly for every 1" > 0, S,(1) =] 1 — r, 1 + Y [ contains a point of A

different from "1". -

Thus ”1" is a limit point of A. '

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.

Let R be the metric space.,Let A = {xix e R, 0 < x < 1 } be a subset

of R. Show that “O” and "1" are the limit point of A.

Solution V

‘

Here metric space is (R, d), where d: R >< R —> R be defined by

iHere A={x|xeR, 0<;c<1}

i =1<>.1[(i) First we shall prove that "O", is the limit point of A. ,

Now Sr(0) = {xlx e_R, ti(x,O) < r}; r._> O

A ‘ ‘ \., | _,

' - .'={x|xeR,-lx—0|<r} -~

= {xlx E R, |x\ < r} , .‘ N,/7

={xlxeR,_x<r, x>-r}\

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1 Clearly for every 1" > O, ST(0) =] -1", _+r [contains a point of A

different from 0 . .

Thus "O" is ia limit point of A.

. (ii) Now we shall prove that "1" isthe limit point of A.

-_ Now .S',.(1) = {xlx e R, d(x, 1) < 1"} -p A h '1l

I

, I

i.

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={x|xER, |xp—1l<r} .-

.

i

i={x|x_eR,x<A1+r,x>1—r}.={x|xeR,1—r<x<_1+r}=]1_T',1+T[ ~ A

- ' I,

A Clearly for everypr > O, AS,(1) =i] 1 — r, ll + r [ contains a point of A

A different from "1". ,1 A AA

- Thus ”1”vis a limit point of A.

Question .

' .

_. Let R be the metric space. Describe the limit points of the followings.

. (a)N (b)Z .

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Here metric space is (R, cl), where d: >< R —> R be defined by A

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A.

(a) Here N = { 1,2,3,-

Let a E R be a limit point of N.

Then aeN or a€EN

Case-I When a e N

Then $r(a) = {x|x e R, d(x, a) < 1-}; r > O

=,[x|x‘e R, |x4—a| <1}={x|xeR, x—a<Ir, )¢—a>—-r}={x|xeR, x<a+r, )¢'>a—r]_"-f{x|xeR, a-—r<Ix< a+r}=]a—r, a+ r[

Clearly for every r > 0, S,(1) =] a — r, a + r [ contains ‘no[ll l

II III point of N different from a .

Thus ”a” is not the limit point-of N.

Case - II When a € N, we can also prove that "4" is not a limit point of N.I Thus N has no limit point. '

| \

53.‘. I (b) .Here Z ={ .... ..— 3,—2,-—1,0,]'.,2,.. I

Let a e R be a limit point of Z.

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Then aEZ or a6£Z

"Case-I Whena e Z

Then S,(a) = {xlx e R, d(_X.a) < 1"}; r > 0

={x|xeR,|x—a|<1*'}={x|xeR, x——a<r, x-a>—r}={x|xeR, x<a+r, x7a—r}={x|xeR, a—r<x<-ia I-r} -

= a—r,a+r[.

Clearly for every r > O, S,_(1) =]l a -r , 'a + r [ contains no

\\

\\

point of Z different from a . “

Thus "a" is not the limit point of Z.

Case — II When a E Z, we can also prove that "'a” IS not a limit point of Z.

Thus Z has no limit point.~

I

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Example T

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n -" T -

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i

=> ] - r , 1" [~ is a neighbourhood of "IO". -

l\lOWOE]-T, r[<_Z.]—r, 1"] ‘Where]—T, 1"[isanopen

ii; ,. => ] - 1", 1"] is a neighbourhood of "O”.' N

i» (ill) NOWOE]—T, T[§[—-T, T[ Where]—-1", T[isanopen

l

= [-1" , 1" [ is a neighbourhood of ”O”. »

1

(iv) NOW O E] -7‘, T [ Q[—T, T] Where] — T, 7"[ is anopensphere in R‘

L’/V.‘ => [-1" , r] is a neighbourhood of

Theorem ~_

F \° _ ‘Let (X, d) be a metric space. LetA Q X: Let x,, .be' a limit point of A. .

Then every neighbourhood of x, contains infinitely many points of A. -»

Proof ' ‘ , _

- '

.

' 1

Let N be a neighbourhood of x, , then 3 an open sphere ST (-x,,) .

(where 1" > 0 ) such that l ‘

_, nE$UJ;N --------------»m

We are to prove that N contains infinite points o§__A_,,,____..;- 1

We prove it by contradiction method. ' V L

\ F '5

Suppose N contains finite points of A. h

Then by (1) Sr (\x;,) also contains finite points of A. T T 1/ T

Suppose S,(x,,) containsn points x1,x2,x3, .... ..,xn of A.

Then ' A O S,(x.,) = {x1,x?_,x3, ...,xn} i

Let d(x.,,xi) = 1",, i = 1,2,3, .... ..,n

" Let r’ = min (1"1,1"2,1"3, ..;,1",,)I

<==::"i'" 1~ -r-----—-—~.—-~ 1- »_*—,~~..--- —

neighbourhood of x,,, if E! an open sphere ST (xo) such thatx, e S, (xo). Q N.

1

Let R-be the usual metric space. Let x., = 0 E R. Show that] 4 1', r[, - ’ ‘,}

h L] - r, r], [-1", r[, and [—r, r],(r > 0 ) is a neighbourhood of 0.

In a usual metric space R, the open sphere is an open interval.

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